Äîêóìåíò âçÿò èç êýøà ïîèñêîâîé ìàøèíû. Àäðåñ îðèãèíàëüíîãî äîêóìåíòà : http://theory.sinp.msu.ru/~smirnov/PSI08.pdf Äàòà èçìåíåíèÿ: Tue Nov 4 18:43:49 2008 Äàòà èíäåêñèðîâàíèÿ: Mon Oct 1 20:02:57 2012 Êîäèðîâêà: Ïîèñêîâûå ñëîâà: m 44

New Methods for Feynman Integrals
V.A. Smirnov Nuclear Physics Institute of Moscow State University

V.A. Smir nov

PSI, November 03, 2008 ­ p.1


Introduction. Methods of evaluating Feynman integrals Reduction to master integrals using IBP relations. Lapor ta algorithm and its implementations. FIRE Evaluating Feynman integrals by sector decompositions. FIESTA Evaluating Feynman integrals by the method of Mellin­Barnes representation. MBresolve.m A recent application: the three-loop quark static potential

V.A. Smirnov

PSI, November 03, 2008 ­ p.2


A given Feynman graph tensor reduction various scalar Feynman integrals that have the same structure of the integrand with various distributions of powers of propagators. dd k1 dd k2 ... (p2 - m2 )a1 (p2 - m2 )a2 ... 1 1 2 2

F (a1 ,a2 ,...) = d=4-2

...

Besides usual propagators, one can have
1 (v · k + i0)
V.A. Smirnov

a

PSI, November 03, 2008 ­ p.3


Methods to evaluate Feynman integrals: analytical, numerical, semianalytical . . . An old straightforward analytical strategy: to evaluate, by some methods, every scalar Feynman integral generated by the given graph.

V.A. Smirnov

PSI, November 03, 2008 ­ p.4


The standard modern strategy: to derive, without calculation, and then apply IBP identities between the given family of Feynman integrals as recurrence relations. A general integral of the given family is expressed as a linear combination of some basic (master) integrals. The whole problem of evaluation constructing a reduction procedure evaluating master integrals

V.A. Smirnov

PSI, November 03, 2008 ­ p.5


Solving reduction problems algorithmically: `Lapor ta's algorithm'
[S. Lapor ta and E. Remiddi'96; S. Lapor ta'00; T. Gehrmann and E. Remiddi'01]

A public version AIR Private versions
D. Seidel, V. Velizhanin, ...]

[C. Anastasiou and A. Lazopoulos'04]

[T. Gehrmann and E. Remiddi, M. Czakon, Y. SchrÆder, C. Sturm, P. Marquard and

Baikov's method GrÆbner bases. Suggested by O.V. Tarasov An alternative approach:
[A.V. Smirnov & V.A. Smirnov'05­07; A.G. Grozin, A.V. Smirnov and V.A. Smirnov'06 A.V. Smirnov, V.A. Smirnov, and M. Steinhauser'08 ]
V.A. Smirnov PSI, November 03, 2008 ­ p.6

[O.V. Tarasov'98]


FIRE = Feynman Integrals REduction (implemented in Mathematica)

[A.V. Smirnov'08]

http://www-ttp.par ticle.uni-karlsruhe.de/asmirnov Sectors 2n regions labelled by subsets {1,... ,n}: = {(a1 ,... ,an ) : ai > 0 if i , ai 0 if i } Natural ordering. The goal of reduction: to make more non-positive indices.

V.A. Smirnov

PSI, November 03, 2008 ­ p.7


Three different strategies in FIRE. 1. In sectors with a small number of non-positive indices, apply s-bases (generalizations of GrÆbner bases). Constructing them automatically by a kind of Buchberger algorithm. SBases.m

V.A. Smirnov

PSI, November 03, 2008 ­ p.8


2. In sectors with a large number of non-positive indices, integrate over a loop momentum explicitly and reduce the problem to a family of two-loop integrals where the index of one propagator is, possibly, shifted by or 2 . Consider the region a2 ,a5 ,a10 0 , a7 ,a8 > 0
a10 a9 a2 a11 a6 a1 a7 a8 a3 a2 a6 a4 a1 a5 + a4 a7 a3

a5

Apply s-bases (within FIRE) to such 2loop reduction problems with 7 indices.
V.A. Smirnov PSI, November 03, 2008 ­ p.9


Reduce indices a2 ,a5 ,a10 ,a7 ,a8 to their boundary values, i.e. a2 ,a5 ,a10 = 0, a7 ,a8 = 1
a10 a2 0 0

a7

a8



1

1

a5

0

At these values, the transition to the 2loop problem because very simple (without multiple summations).

V.A. Smirnov

PSI, November 03, 2008 ­ p.10


3. In `intermediate sectors', the Lapor ta's algorithm (implemented within FIRE) is applied. FIRE can be run in a `pure Lapor ta' mode. `Lee ideas' [R.N. Lee'08] In each sector one may find a single IBP that works for `most' points in this sector. One might generate less IBPs because the other IBPs are naturally represented as a linear combination of these. QLink is used to access the QDBM database for storing data on disk from Mathematica. FLink allows to perform external evaluations by means of the Fermat program. Fermat speds up Together and GCD. http://www-ttp.par ticle.uni-karlsruhe.de/asmirnov
V.A. Smirnov PSI, November 03, 2008 ­ p.11


Methods to evaluate master integrals: Feynman/alpha parameters Mellin­Barnes representation
T. Gehrmann & E. Remiddi'00] [V.A. Smirnov'99, J.B Tausk'99] [A.V. Kotikov'91, E. Remiddi'97,

method of differential equations

V.A. Smirnov

PSI, November 03, 2008 ­ p.12


UV, IR and collinear divergences Regularization. Dimensional regularization. Formally, d4 k = dk0 k dd k where d = 4 - 2 Informally, use alpha parameters
1 ia = 2 + m2 - i0)a (-k (a)
0

d

a-1 i(k 2 -m2 )

e

change the order of integration, take Gauss integrals over the loop momenta

V.A. Smirnov

PSI, November 03, 2008 ­ p.13


d ke

d

4

i(k 2 -2q ·k )

= -i

2 -2 -iq 2 /

e

d ke

i(k 2 -2q ·k )

=e

i (1-d/2)/2 d/2 -d/2 -iq 2 /





e

V.A. Smirnov

PSI, November 03, 2008 ­ p.14


Graph
ia+
h(1-d/2) hd/2 l

F (a1 ... ,aL ; d) = â
0

(al )
l al -1 l

d1 ...

0

d

L

U

-d/2 iV /U-i

e

m2 l l

,

where h is the number of loops and
U=
trees T lT

l ,
2-

V=
trees T lT
V.A. Smirnov

l

q

T

2

.
PSI, November 03, 2008 ­ p.15


F (q1 ,... ,qn ; d) = â
0

i

d/2

h

(a - hd/2) (al ) U
a-(h+1)d/2 al -1 ll 2 a-hd/2 ll

l

d1 ...

0

dL

l - 1



-V + U

m

Sector decompositions. Hepp sectors
1 2 ...
L

[K. Hepp'66]

sector variables tl = l / Back:
V.A. Smirnov

l+1

, l = 1,... ,L - 1; tL = L .

l = tl ...tL
PSI, November 03, 2008 ­ p.16


Speer's sectors labelled by the elements of a UV forest F ,
l =
F : l

[E. Speer'77]

t

For Feynman integrals with the Euclidean external momenta (( qi )2 < 0 for any subset of external momenta), Speer's sectors are optimal for the resolution of the singularities of the integrand.

V.A. Smirnov

PSI, November 03, 2008 ­ p.17


Recursively defined sector decompositions
[T. Binoth and G. Heinrich'00]

Primary sectors
i l , l = i = 1, 2,... ,L ,

with new variables
ti = i / l
l

if i = l if i = l

The contribution of a primary sector 1 1 dti U Fl = ...
V.A. Smirnov

L-(h+1)d/2

0

0

i =l

V

L-hd/2

tl =1
PSI, November 03, 2008 ­ p.18


Next sectors are introduced in similar way. The goal is to obtain a factorization of U and V in final sector variables, i.e. to represent them as products of sector variables in some powers times a positive function. Strategies that are guaranteed to terminate
[C. Bogner & S. Weinzierl'07]

A, B, C, X
[A.V. Smirnov & M.N. Tentyukov'08] Strategy S FIESTA (Feynman Integral Evaluation by a Sector decomposiTion Approach) http://www-ttp.par ticle.uni-karlsruhe.de/asmirnov

V.A. Smirnov

PSI, November 03, 2008 ­ p.19


The usage of Speer's sectors within FIESTA. It turns out that, for Feynman integrals at Euclidean external momenta, Speer' sectors are reproduced within Strategy S
[A.V. Smirnov & V.A. Smirnov'08]

(e.g. for four-loop propagator integrals) m62: 26304 sectors

V.A. Smirnov

PSI, November 03, 2008 ­ p.20


'$

m61

'$ &%

m62

'$ $$$ &%

m63

&%

'$ ¡ ¡ rr ¡r ¡ &%

m51

'$ t t t t &%

m41

'$ t e t e t t &%

m42

'$ &%

m44

'$ t t t t &%

m45

'$ e ¡ e ¡ e¡ e¡ &%

m34

'$ e e ¡ ¡ &%

m35

'$

m36

m52
d d d

&%

'$ d d &%

m43

m31

m32


'$ &%

m33

'$ '$ r rr &% &%

m21

'$ d d d d &%

m22

m23

m24
d d

'$ ¢ f f ¢ &%

m25

'$ &%

m26

'$ '$

m27

m11


&% &%

m12


'$ &%

m13

'$ &%

m14



m01

Again, as in 3­lo op case, "glue­and­cut" relations provide with enough information to express co efficient of expansion over e = 2 - D/2 of all these integrals

8


MB
Mellin transformation, Mellin integrals as a tool for Feynman [M.C. BergÕre & Y.-M.P. Lam'74] integrals: Evaluating individual Feynman integrals:
[N.I. Ussyukina'75..., A.I. Davydychev'89...,]

Systematic evaluation of dimensionally regularized Feynman integrals (in par ticular, systematic resolution of [V.A. Smirnov'99, J.B. Tausk'99] the singularities in )

V.A. Smirnov

PSI, November 03, 2008 ­ p.21


The basic formula:
1 11 = () 2 i (X + Y )
+i -i

Y dz X

z +z

(+z )(-z ) .

The poles with a (... +z ) dependence are to the left of the contour and the poles with a (... -z ) dependence are to the right

V.A. Smirnov

PSI, November 03, 2008 ­ p.22


C

2

Im z

- - 2

- - 1

1 - 0 1 Re z 2

-2

-1

-1

-2
V.A. Smirnov PSI, November 03, 2008 ­ p.23


The simplest possibility:
11 1 = 2 - k 2 ) () 2 i (m
+i -i

( m2 ) z dz ( + z )(-z ) 2 ) + z (-k

Example 1

F (q 2 ,m2 ; a1 ,a2 ,d) =

dd k (m2 - k 2 )a1 (-(q - k )2 )

a2

V.A. Smirnov

PSI, November 03, 2008 ­ p.24


dd k = i 2 )a1 [-(q - k )2 ]a2 (-k

d/2

G(a1 ,a2 ) (-q 2 )a1 +a2 +

-2

,

(a1 + a2 + - 2)(2 - - a1 )(2 - - a2 ) G(a1 ,a2 ) = (a1 )(a2 )(4 - a1 - a2 - 2 ) i d/2 (-1)a1 +a2 (2 - - a2 ) F (q 2 ,m2 ; a1 ,a2 ,d) = (a1 )(a2 )(-q 2 )a1 +a2 + -2 1 m â dz (a1 + a2 + - 2+ z ) 2 2 i -i -q (2 - - a1 - z )(-z ) â (4 - 2 - a1 - a2 - z )
+i 2 z

V.A. Smirnov

PSI, November 03, 2008 ­ p.25


F (q ,m ;1, 1,d) = 1 â 2 i

2

2

i
2 2

d/2

(1 - ) (-q 2 ) ( + z )(-z )(1 - - z ) (2 - 2 - z )

dz
C

m -q

z

( +z )(-z ) a singularity in

V.A. Smirnov

PSI, November 03, 2008 ­ p.26


C

C

2

Im z

1 - - 1 -2 -1 - 0 1- 1 Re z 2

-1

V.A. Smirnov

-2

PSI, November 03, 2008 ­ p.27


C, C

2

Im z

1 - - 2 -2 - - 1 -1 C 0

C -

1- 1 Re z 2

-1

V.A. Smirnov

-2

PSI, November 03, 2008 ­ p.28


Take a residue at z = - :
( ) i (m2 ) (1 - )
2

and shift the contour:
1 i 2 i
2

dz
C

m -q

2 2

z

(z )(-z ) 1-z

V.A. Smirnov

PSI, November 03, 2008 ­ p.29


Example 2. The massless on-shell box diagram, i.e. with p2 = 0, i = 1, 2, 3, 4 i
p
1

1 2 4 3

p

3

p

2

p

4

F (s, t; a1 ,a2 ,a3 ,a4 ,d) =

dd k , 2 )a1 [(k + p )2 ]a2 [(k + p + p )2 ]a3 [(k - p )2 ]a4 (k 1 1 2 3
2

where s = (p1 + p2 )2 and t = (p1 + p3 )
V.A. Smirnov

PSI, November 03, 2008 ­ p.30


F (s, t; a1 ,a2 ,a3 ,a4 ,d) a d/2 (a + - 2)(2 - - a1 - a2 )(2 - - a3 - a4 ) = (-1) i (4 - 2 - a) (al ) â
1 0 0 1 a a 1 1 -1 (1 - 1 )a2 -1 2 3 -1 (1 - 2 )a4 - [-s1 2 - t(1 - 1 )(1 - 2 ) - i0]a+ 1 -2

d1 d2 ,

where a = a1 + a2 + a3 + a4 Apply the basic formula to separate -s1 2 and -t(1 - 1 )(1 - 2 ) in the denominator Change the order of integration over z and -parameters, evaluate parametric integrals in terms of gamma functions
V.A. Smirnov PSI, November 03, 2008 ­ p.31


(-1)a i d/2 F (s, t; a1 ,a2 ,a3 ,a4 ,d) = (4 - 2 - a) (al )(-s)
z

a+ -2

+i 1 t â dz (a + - 2+ z )(a2 + z )(a4 + z )(-z ) 2 i -i s â(2 - a1 - a2 - a4 - - z )(2 - a2 - a3 - a4 - - z )

V.A. Smirnov

PSI, November 03, 2008 ­ p.32


General prescriptions
Derive a (multiple) MB representation for general powers of the propagators. (The number of MB integrations can be large (more than 10)). Use it for checks. Reducing a line to a point tending ai to zero (usually) taking some residues. A typical situation: (a2 +z )(-z ) , a2 0 (a2 ) Gluing of poles of different nature. Take a (minus) residue at z2 = 0, then set a2 = 0. Unambiguous prescriptions for choosing integration contours Try to have a minimal number of MB integrations.

V.A. Smirnov

PSI, November 03, 2008 ­ p.33


Resolve the singularity structure in . The goal: to represent a given MB integral as a sum of integrals where a Laurent expansion in becomes possible. The basic procedure: take residues and shift contours

V.A. Smirnov

PSI, November 03, 2008 ­ p.34


Two strategies: #1 [V.A. Smirnov'99 ] E.g., the product (1+z )(-1 - -z ) generates a pole of the type (- ). The general rule: (a+z )(b-z ), where a and b depend on the rest of the variables, generates a pole of the type (a + b). Analysis of the integrand. Analysis of integrations over MB variables in various order leads to understanding what `key' gamma functions (responsible for the generation of poles in ) are. `Changing the nature' of these gamma functions (i.e. changing rules for the contours). Taking residues and shifting contours.
V.A. Smirnov PSI, November 03, 2008 ­ p.35


#2

[J.B. Tausk'99, Anastasiou'05, Czakon'05 ]

.

Choose a domain of and Rezi ,...Rewi in such a way that all the integrations over the MB variables can be performed over straight lines parallel to imaginary axis. Let 0. Whenever a pole of some gamma function is crossed, take into account the corresponding residue. For every resulting residue, which involves one integration less, apply a similar procedure, etc. Two algorithmic descriptions
[C. Anastasiou'05, M. Czakon'05 ]

The Czakon's version MB.m implemented in Mathematica is public.

V.A. Smirnov

PSI, November 03, 2008 ­ p.36


#1 in a slightly modified form Let
ai ( )+
i j

[A.V. Smirnov & V.A. Smirnov'08 ]

bij ( )zj

be the numerator of a multiple MB integral. Let be real. `Changing the nature' of the key gamma functions (i.e. changing rules for the contours)
ai ( )+
j

bij ( )z

j

(1) ai ( )+

j bij

( )z

j



... (n) (...) ... instead of ai ( )+ j bij ( )Rezj > 0 we have -1 < ai ( )+ j bij ( )Rezj < 0,... , -n - 1 < ai ( )+ j bij ( )Rezj < -n,...
V.A. Smirnov PSI, November 03, 2008 ­ p.37


Strategy #2: straight contours in the beginning Strategy #1: straight contours in the end Set = 0 Look for straight contours (i.e. Rezi ) for which gamma functions are changed in a minimal way.

V.A. Smirnov

PSI, November 03, 2008 ­ p.38


Let S (x) = [(1 - x)+ ] where [...] is the integer par t of a number and x+ = x for x > 0 and 0 otherwise. Look for contours for which i S ai (0) + j bij (0)zj is minimal. With such a choice, identify gamma functions which should be `changed'. Take a residue and replace by (1) (and, possibly, (1) by (2) etc.) Proceed iteratively. MBresolve.m http://www-ttp.par ticle.uni-karlsruhe.de/asmirnov

V.A. Smirnov

PSI, November 03, 2008 ­ p.39


Evaluate MB integrals after expanding in . Apply the first and the second Barnes lemmas
+i 1 dz (1 + z )(2 + z )(3 - z )(4 - z ) 2 i -i (1 + 3 )(1 + 4 )(2 + 3 )(2 + 4 ) = (1 + 2 + 3 + 4 )

1 2 i

V.A. Smirnov

(1 + z )(2 + z )(3 dz (6 -i (1 + 4 )(2 + 4 )(3 + = (1 + 2 + 4 + 5 )(1 + (2 + 5 )(3 + 5 ) , 6 = â (2 + 3 + 4 + 5 )

+i

+ z )(4 - z )(5 - z ) + z) 4 )(1 + 5 ) 3 + 4 + 5 ) 1 + 2 + 3 + 4 +
5

PSI, November 03, 2008 ­ p.40


Example 3. Non-planar two-loop massless ver tex diagram with p2 = p2 = 0, Q2 = -(p1 - p2 )2 = 2p1 · p2 1 2
p 3 q 1 2 4 p
2 1

5 6

F (Q2 ; a1 ,... ,a6 ,d) =

dd k dd l [(k + l)2 - 2p1 · (k + l)]a1
a6

1 â [(k + l)2 - 2p2 · (k + l)]a2 (k 2 - 2p1 · k )a3 (l2 - 2p2 · l)a4 (k 2 )a5 (l2 )
V.A. Smirnov

PSI, November 03, 2008 ­ p.41


Gonsalves'83:
F (Q2 ; a1 ,... ,a6 ,d) = â(a +2 - 4)
a â3 1 -1 a2 -1 4 1 0

(-1)
1

a

i

d/2

2

(2 - - a35 )(2 - - a46 ) (al ) (4 - 2 - a3456 )
a5 -1 a4 -1 2 4 -2 -a a6 -1

(Q2 )
0

a+2 -4 -1

d1 ...

a d4 1 3

(1 - 1 )

(1 - 2 )

(1 - 3 - 4 )

a3456 + -3 +

A(1 ,2 ,3 ,4 )

,

where
A(1 ,2 ,3 ,4 ) = 3 4 +(1 - 3 - 4 )[2 3 (1 - 1 )+ 1 4 (1 - 2 )]

V.A. Smirnov

PSI, November 03, 2008 ­ p.42


[ (1 - )+(1 - )(2 (1 - 1 )+ (1 - )1 (1 - 2 ))]a+2

(a +2 - 4)

-4

dz1 (-z1 ) z1 z1 (1 - )z1 (1 - )a+2 -4+z1 -i (a +2 - 4+ z1 ) â [2 (1 - 1 )+(1 - )1 (1 - 2 )]a+2 -
1 = 2 i

+i

4+z1

The last line
1 2 i
V.A. Smirnov

+i -i

z dz2 (a +2 - 4+ z1 + z2 )(-z2 ) z2 22 (1 - 1 )z2 a (1 - )a+2 -4+z1 +z2 1 +2 -4+z1 +z2 (1 - 2 )a+2 -4+z1 +

z2

PSI, November 03, 2008 ­ p.43


F (Q2 ; a1 ,... ,a6 ,d) =

(-1) (Q2 )

a

i

d/2

2

(2 - - a35 ) (al )

a+2 -4

(6 - 3 - a)

+i +i 1 (2 - - a46 ) dz1 dz2 (a +2 - 4+ z1 + z2 ) â 2 (4 - 2 - a3456 ) (2 i) -i -i â(-z1 )(-z2 )(a4 + z2 )(a5 + z2 )(a1 + z1 + z2 ) (2 - - a12 - z1 )(4 - 2 + a2 - a - z2 ) â (4 - 2 - a1235 - z1 )(4 - 2 - a1246 - z1 ) â(4 - 2 + a3 - a - z1 - z2 )(4 - 2 + a6 - a - z1 - z2 ) ,

where a

3456

= a3 + a4 + a5 + a6 , etc.

V.A. Smirnov

PSI, November 03, 2008 ­ p.44


In[1]:= In[2]:=

SetDirectory "c: diskE job2008 Bern" ; MB MB.m; MB MBresolve.m

MB 1.1 by Michal Czakon more info in hep ph 0511200 last modified 06 Mar 08 MBresolve 1.0 by Alexander Smirnov last modified 22 Oct 08 2fold MB representation for the non planar vertex massless diagram. The factor QQ4 a1 a2 a3 a4 a5 a6 2 ep is omitted . QQ p1 p2 ^2. The factor I Pi^ d 2 ^2 is also omitted as usually.
In[4]:=

NPMB a1 , a2 Gamma Gamma 2 Gamma 6 Gamma a1 Gamma 2 Gamma 4 Gamma a5 Gamma

, a3 , a4 , a5 , a6 : 1 ^ a1 a2 a3 a1 Gamma a2 Gamma a3 Gamma a4 Gamma a5 ep a3 a5 Gamma 2 ep a4 a6 Gamma 4 3 ep a1 a2 a3 a4 a5 a6 a2 a3 a4 a5 a6 2 ep 4 z1 z2 Gamma ep a1 a2 z1 Gamma a4 z2 Gamma a1 z1 2 ep a1 a3 a4 a5 a6 z2 Gamma 4 2 ep z2 Gamma 4 2 ep a1 a2 a3 a4 a5 z1 4 2 ep a1 a2 a4 a6 z1 Gamma 4 2 ep

a4 a5 a6 Gamma a6 2 ep a3 a4

a5

a6

z1 Gamma z2 z2 a1 a2 a4 a5 a6 z1 z2 a1 a2 a3 a5 z1 ;

z2

The diagram with all powers of the propagators equal to one. We shall evaluate it in expansion in ep up to ep^0.
In[5]:= Out[5]=

V2

NPMB 1, 1, 1, 1, 1, 1 ep
2

Gamma Gamma Gamma

Gamma

ep

z1 Gamma z2
2

z1 Gamma z1
2

1

2 ep

z2 Gamma 2 ep z1 z2

1

2 ep

z1

z2

2

z2 Gamma 1 3 ep Gamma

Gamma 1

z2 Gamma 2 z1

2 ep Gamma

2 ep 0,

In[6]:=

V2rules 5 ep 8 V2cont

MBoptimizedRules V2, ep 1 , z1 4 , z2 4 1

, ep

Out[6]=

In[7]:=

MBcontinue V2, ep

0, V2rules ;


2

Ex3.nb

Level 1 Taking Taking Level 2 Integral Taking Integral Level 3 Integral 1, 1 found MBpreselect MBmerge V2cont , ep, 0, 0 1 residue in z1 2 2 ep residue in z2 residue in z2 1 1 2 ep 2 ep z1

4 integral s
In[8]:= In[9]:=

V2select V2exp MBint ep MBint
4

Simplify MBexpand V2select, Exp 2 ep EulerGamma , ep, 0, 0 1 2 2 ep
2

41 4 40 z1
2

55 PolyGamma 2, 1 , 3 ep ep 0, ,

Out[9]=

1 Gamma 4 ep 12
2 2

Gamma z1 Gamma 1 6 ep2 EulerGamma
2

z1 7 ep2 2 6 ep2 PolyGamma 0, z1
2

12 ep EulerGamma

12 ep 1 3 66 ep ep

PolyGamma 0, z1 2 12 ep2 PolyGamma 0, 1 z1 2 24 ep PolyGamma 0, z1 ep EulerGamma ep PolyGamma 0, 1 z1 12 ep PolyGamma 0, z1 3 ep EulerGamma 2 ep PolyGamma 0, z1 4 ep PolyGamma 0, 1 z1 2 PolyGamma 1, z1 12 ep2 PolyGamma 1, z1 12 ep2 PolyGamma 1, 1 z1 1 , MBint 6 Gamma 4
2

,

0 , z1 z2

1

z2 Gamma z1 z2 ,

1 ep

z1

z2

2

Gamma 1

z2 1

Gamma 1
In[10]:= In[11]:=

Gamma 1

z1

z2 Gamma 2

0 , z1 4

, z2 4

MBresolve V2, ep Simplify Gamma 2 ep
4

Gamma Gamma

ep

2

Gamma 1
2

2 ep

2

Out[11]=

MBint 4 ep ep
2

, Gamma 2 ep Gamma z1 Gamma 1

ep

0, ep
2

, z1
3

MBint

2 ep Gamma 2 ep z1

z1 Gamma ep

Gamma 1 0.859981

z1 ,

Gamma MBint

2 ep 1

Gamma ep
2

3 ep Gamma Gamma 1

,

0 , z1 ep

Gamma

2 ep

z2 Gamma 2 ep z2 1

z2 Gamma 1

z2 z2 0.859981 , z2
2

Gamma 3 ep Gamma 1 2 ep z2 PolyGamma 0, 1 MBint Gamma ep
2

EulerGamma ep z2 ep z2
2

PolyGamma 0, 2 ep

2 PolyGamma 0, 1 , ep 0 , z2 z2 Gamma z1 z2 1

PolyGamma 0, 1 z1 Gamma Gamma 1 z1

Gamma

z1 Gamma
2

2 ep 2 ep

2 ep

z1

Gamma Gamma ep
In[12]:=

z2 Gamma 1 3 ep Gamma

z2 Gamma 2 z1 ,

2 ep Gamma

2 ep

0 , z1

0.72274, z2

0.274294 , ep, 0, 0

V2select

MBpreselect MBmerge


Ex3.nb

3

In[13]:=

V2expS MBint

Simplify MBexpand 1 2
4

, Exp 2 ep EulerGamma , ep, 0, 0 55 PolyGamma 2, 1 , ep 0, , 3 ep

41 4
2

Out[13]=

ep MBint

2 ep

40 z1
2

1 Gamma 8 ep 12
2

Gamma z1 Gamma 1 6 ep2 EulerGamma
2

z1 ep2 2 54 ep2 PolyGamma 0,
2

12 ep EulerGamma

z1

2

24 ep 1 ep EulerGamma PolyGamma 0, z1 36 ep PolyGamma 0, z1 1 ep EulerGamma 42 ep2 PolyGamma 1, 1 MBint 8 ep 4
2

24 ep2 PolyGamma 0, z1 2 ep PolyGamma 0, z1 , ep 0 , z1

z1

24 ep2 PolyGamma 1, z1 z2 Gamma 1
2

0.859981

,

3 Gamma

1

z2 Gamma

z2

2

4 ep EulerGamma

2 ep2 EulerGamma

5 ep2 2

8 ep2 PolyGamma 0,

1

z2

2 2

12 ep 1 ep EulerGamma PolyGamma 0, 1 z2 8 ep PolyGamma 0, 1 z2 1 ep EulerGamma 8 ep2 PolyGamma 1, MBint 6 Gamma Gamma 2 z1 1 1 z2 1

18 ep2 PolyGamma 0, 1 z2 3 ep PolyGamma 0, 1 z2 z2 , ep z2
2

14 ep2 PolyGamma 1, 1 z1 z2
2

0 , z2 Gamma 1

0.859981 z1 z2

,

z2 Gamma ep

Gamma

z2 Gamma 1 0.274294

z2 ,

0 , z1

0.72274, z2

In[14]:=

res1 1

V2exp 2

1 41 4

1 55 PolyGamma 2, 1 3 ep

Out[14]=

ep
In[15]:= Out[15]=

4

2 ep 3

2

40

V2exp

MBint 6 Gamma Gamma 1 z1

1

z2 Gamma

1 z1

z1

z2

2

Gamma

z2 Gamma 1 1 , z2 4

z2

2

1 4

z2 Gamma 2 3 1 , z1 z2 Gamma 1

z2 ,

ep

0 , z1

In[16]:=

Barnes1 V2exp MBint 2 Gamma res2 4 Barnes1

Out[16]=

z2 Gamma 1

z2

2

1 , ep 0 , z2 4

In[17]:=

, z2

Out[17]=

6
In[18]:=

V2exp MBint

2 1

Out[18]=

4 ep Gamma
2

2 2

z1

Gamma z1 Gamma 1
2

z1
2

12

12 ep EulerGamma
2

6 ep2 EulerGamma
2

2

7 ep2 2

6 ep PolyGamma 24 ep PolyGamma PolyGamma 0, 66 ep2 PolyGamma

0, z1 12 ep PolyGamma 0, z1 1 ep EulerGamma z1 3 3 ep EulerGamma 2 1, z1 12 ep2 PolyGamma z1 1 , ep

12 ep2 PolyGamma 1, 1
In[19]:=

0, z1 12 ep PolyGamma 0, 1 z1 2 ep PolyGamma 0, 1 z1 12 ep ep PolyGamma 0, z1 4 ep PolyGamma 0, 1 1, z1 1 0 , z1 4 1 , ep, n ;

z1

CoeffEps X , n

:

X

.X

Simplify Coefficient X


4

Ex3.nb

In[20]:=

CoeffEps V2exp MBint 3 Gamma res32 2

2 z1
2

,

2 1 z1 , 2 , z1 ep 1 0 , z1 4

Out[20]=

Gamma z1 Gamma 1 2 ,

In[21]:=

Barnes1 CoeffEps V2exp

Out[21]=

2
In[22]:= Out[22]=

CoeffEps V2exp MBint 3 Gamma EulerGamma

2 z1
2

,

1 z1 1 , ep 0 , z1 4

Gamma z1 Gamma 1 z1

3 PolyGamma 0,

2 PolyGamma 0, z1

In[23]:= In[24]:= Out[24]=

res31 res31

9 Zeta 3 ; N

10.8185 1

In[25]:=

NIntegrate CoeffEps V2exp y1, Infinity , Infinity 2.13163 10 2
14

2

,

1

1

2 Pi

. z1 4

I y1 ,

Out[25]= In[26]:=

10.8185

CoeffEps V2exp 1 MBint

,0

Out[26]=

Gamma z1 2 Gamma z1 Gamma 1 z1 4 6 EulerGamma2 7 2 6 PolyGamma 0, z1

2

12 PolyGamma 0, z1

2

12 PolyGamma 0, 1 z1

z1

2

24 PolyGamma 0, z1 EulerGamma PolyGamma 0, 1 z1 3 EulerGamma 2 PolyGamma 0, z1 4 PolyGamma 0, 1 12 PolyGamma 1, z1 7
In[27]:=
4

12 PolyGamma 1, 1

z1

,

ep

12 PolyGamma 0, z1 z1 66 PolyGamma 1, 1 0 , z1 4

res30 10 res30 68.1864 N

;

In[28]:= Out[28]=

1
In[29]:=

NIntegrate CoeffEps V2exp y1, Infinity , Infinity 0. result

2

,0

1

2 Pi

. z1 4

I y1 ,

Out[29]=

68.1864

In[30]:=

FullSimplify res1 1 2
4

res2

res32 ep ^ 2

res31 ep

res30

59 4 120

83 Zeta 3 3 ep

Out[30]=

ep

ep

2


In[1]:= In[2]:=

SetDirectory "c: diskE job2008 Bern" ; MB MB.m; MB MBresolve.m

MB 1.1 by Michal Czakon more info in hep ph 0511200 last modified 06 Mar 08 MBresolve 1.0 by Alexander Smirnov last modified 22 Oct 08
In[4]:=

K2 a1 , a2 , a3 , a4 , a5 , a6 , a7 , a8 : xz1 Gamma z1 Gamma a2 z1 Gamma 2 a5 a6 Gamma Gamma Gamma Gamma

a7

ep

z1

z2 z4

2 a1 a2 a8 ep z2 Gamma 2 a4 a5 a7 ep z1 z3 2 a2 a3 a8 ep z3 Gamma a7 z1 z4 Gamma 2 a4 a5 a6 a7 ep z1 a8 z2 z3 z4 Gamma z1 z2 z3 z4 Gamma 2 a1 a2 a3 a8 ep z4 z2 z4 Gamma z3 z4 Gamma a5 z1 z2 z3 z4 a6 a7 2 ep

Gamma a2 Gamma a4 Gamma a5 Gamma a6 Gamma a7 Gamma 4 a4 a5 Gamma 4 a1 a2 a3 a8 2 ep z1 z4 Gamma a8 z1 z2 z3 z4 Gamma a3 z2 z4 Gamma a1 z3 z4
In[5]:= Out[5]=

B2 x
z1

K2 1, 1, 1, 1, 1, 1, 1,

1

Gamma z1 Gamma 1 z1 Gamma 1 ep z1 z2 Gamma 1 ep z2 Gamma 1 ep z1 z3 Gamma 1 ep z3 Gamma 1 z1 Gamma 2 ep z1 z4 Gamma 1 z2 z3 z4 Gamma z1 z2 z3 z4 Gamma ep z4 Gamma z2 z4 Gamma z3 z4 Gamma 1 z1 z2 z3 z4 1 z1 z2 z3 z4

z4

Gamma 2 ep Gamma 2 2 ep z1 z4 Gamma Gamma 1 z2 z4 Gamma 1 z3 z4
In[6]:=

MBoptimizedRules B2, ep

0,

, ep

MBrules::norules : no rules could be found to regulate this integral
Out[6]= In[7]:=

MBresolve B2, ep

CREATING RESIDUES LIST0.8594 seconds FAILED TO RESOLVE
Out[7]= In[8]:= In[9]:=

False B2 rul y 128 con1 K2 1, 1, 1, 1, 1, 1, 1, 1 y; 0, 1 , z1 384 6 0, rul ; , z2 96 , y, ep 73 , z3 384 295 , z4 24 19

MBoptimizedRules B2, y 55 107 , ep

Out[9]=

In[10]:=

MBcontinue B2, y


2

Ex4.nb

Level 1 Taking Level 2 Integral Taking Taking Taking Level 3 Integral Integral Taking Taking Integral Taking Level 4 Integral Integral Taking Integral Level 5 Integral 1, 2, 2, 1 found MBmerge; , ep 0, exp1 i, 2 , i, Length exp1 MBmerge; 1, 2, 1 1, 2, 2 residue in z1 1, 3, 1 y 1, 1 1, 2 residue in z1 residue in z3 1, 3 residue in z1 y y 1 y 1 residue in z1 residue in z2 residue in z3 y 1 1 y y residue in z4 1 y z2 z3

9 integral s
In[11]:= In[12]:=

exp1 con2

MBexpand con1, 1, y, 0, 0 Table MBcontinue exp1 i, 1


Ex4.nb

3

In[13]:=

exp2

MBexpand con2, Exp 2 ep EulerGamma , ep, 0, 0 1 3240 1440 ep
4

MBmerge

Out[13]=

MBint 480 ep Log x 1 MBint ep Gamma 2x
z1

2820 ep2 2 ep2 2 z1
2

31 ep4 4

9720 ep3 PolyGamma 2, 1 , y 0, ep 0, ,

6 Gamma

14 ep3 PolyGamma 2, 1 Gamma z1 Gamma 1 z1

z1 Gamma 1 z1 1 ep EulerGamma 4 ep z1 ep PolyGamma 0, 1 z1 Gamma 1 z1 8 ep PolyGamma 0, z1 2 ep PolyGamma 0, z1 1 1 y 0, ep 0 , z1 , MBint Gamma 1 z2 6 ep 1 2 ep EulerGamma ep PolyGamma 0, 1 z2 3 ep 73 y 0, ep 0 , z2 , 96 1 MBint Gamma 1 z3 2 Gamma 1 z3 2 ep 1 2 ep EulerGamma ep PolyGamma 0, 1 z3 3 ep 295 1 y 0, ep 0 , z3 , MBint 384 Gamma 2 z1 2 xz1 Gamma z1 Gamma 1 Gamma z1 Gamma 1 z1 Gamma 1 z1 Gamma z1 1 y 0, ep 0 , z1 6 2 xz1 Gamma z1 Gamma 1 Gamma z1 Gamma 1 z1 Gamma 1 z1 Gamma z1 1 y 0, ep 0 , z1 6

PolyGamma 0, z1 2 ep PolyGamma 0, Gamma z1 3 ep EulerGamma 5 ep PolyGamma 0, 1 z1 ,
2

Gamma 1

z2

2

PolyGamma 0, 2

z2

,

PolyGamma 0, 2 z2 z1

z3

,

z1 Gamma 1 z2 Gamma 1 z2 Gamma 1 Gamma z1 z2 Gamma 1 z1 z2 Gamma 1 z1 z2 Gamma 2 z1 z2 , 73 1 , z2 , MBint 96 Gamma 2 z1 z3 z1 Gamma 1 z3 Gamma 1 z3 Gamma 1 Gamma z1 z3 Gamma 1 z1 z3 Gamma 1 z1 z3 Gamma 2 z1 z3 , 295 , z3 384

z2

z1

z3

In[14]:= Out[14]=

B2 x
z1

K2 1, 1, 1, 1, 1, 1, 1,

1

y

Gamma z1 Gamma 1 z1 Gamma 1 ep z1 z2 Gamma 1 ep y z2 Gamma 1 ep z1 z3 Gamma 1 ep y z3 Gamma 1 Gamma 2 ep z1 z4 Gamma 1 y z2 z3 z4 Gamma z1 z2 z3 z4 Gamma ep y z4 Gamma z2 z4 Gamma z3 z4 Gamma 1 z1 z2 z3 z4 1 y z1 z2 z3 z4

z1

z4

Gamma 2 ep Gamma 2 2 ep y z1 z4 Gamma Gamma 1 z2 z4 Gamma 1 z3 z4
In[15]:= In[16]:= In[17]:= In[18]:=

res1 res2 res3 res4

MBresolve B2, ep

0, y

0

MBpreselect res1, y, 0, 0 MBexpand res2, 1, y, 0, 0 MBpreselect res3, ep, 0, 0


4

Ex4.nb

In[19]:=

res5

Simplify MBexpand res4, Exp 2 ep EulerGamma , ep, 0, 0 5 19 2
4

13 4 720 Log x

Out[19]=

MBint 2 ep 12 ep
2

2 Log x 6

2

2 Log x 3 ep

3

Log x 3

4

47 PolyGamma 2, 1 6 ep 1 MBint ep 1 2x
1 z4

7 ep

2

2

20 ep PolyGamma 2, 1 , 3 ep
3

3

ep

0,

,

Gamma 1

z4

2

Gamma 2

z4 Gamma

1

z4

2

Gamma z4

2 ep PolyGamma 0, 1 z4 ep PolyGamma 0, z4 , 1 ep 0 , z4 1.18859 , MBint 2 x 1 z4 Gamma 1 z4 2 Gamma 2 z4 Gamma 1 ep Gamma z4 1 ep EulerGamma 2 ep PolyGamma 0, 1 z4 ep PolyGamma 0, z4 , ep MBint 0 , z4 1 1.19354 ,
3

ep EulerGamma

z4

2

2 x 1 z4 Gamma 1 z4 ep 4 ep PolyGamma 0, 1 z4 0 , z4 0.649289 z4
2

Gamma

1

z4 Gamma z4

2

1

ep EulerGamma ,

ep

2 ep PolyGamma 0, 1 z4 ep PolyGamma 0, z4 1 , MBint 2 x 1 z4 Gamma 1 z3 Gamma 1 z3 z4 Gamma 1 z3 z4 Gamma z4 , MBint Gamma 1 z2 z4
2 2

Gamma 1 ep 2x
1 z4

z3 Gamma 1

Gamma z3 1

z4

2

,

0 , z3 Gamma 0 , z2 120 1

0.965164, z4 z2 Gamma 1 0.046298, z4 173 ep4 4

0.969365 z2 Gamma 1 0.940037

z4 ,

2

Gamma 1

z2

z4 Gamma z4

Gamma z2

z4

2

,

ep MBint

180 ep2 2

520 ep3 PolyGamma 2, 1 , ep 0, ,
4

480 ep 1 MBint ep 3 ep PolyGamma 0, 2 1 MBint ep 3 ep PolyGamma 0, 2 1 MBint 1 ep z4 Gamma 1 z4
2

Gamma 1

z4

2

Gamma z4 Gamma

1

z4

2

1

2 ep EulerGamma 1 z4 ,

2 ep PolyGamma 0, 1 ep 0 , z4 1.19354

z4 , z4 ,

ep PolyGamma 0, 1 z4
2

1

2 ep EulerGamma 1 z4 ,

2 ep PolyGamma 0, 1 ep 0 , z4 1.19354

ep PolyGamma 0,

2 xz1 Gamma 1 z1 Gamma z1 2 Gamma z1 2 Gamma 1 z1 ep ep EulerGamma 2 ep PolyGamma 0, z1 3 ep PolyGamma 0, 1 0 , z1 0.377479 , MBint 2 x
z1

z1

,

Gamma 1

z1 Gamma 1

z1 Gamma z1 1 z1 z4 , z1 z4

Gamma 1 z1 Gamma 1 z4 Gamma 1 z1 z4 Gamma ep 0 , z1 0.731822, z4 0.76808 , MBint 2 x Gamma
z1

z4 Gamma z1 Gamma 1

Gamma 1

z1 Gamma 1 z1

z1 Gamma z1 Gamma 1 z4 , ep 0 , z1

z4 Gamma 1 0.832684

1

z4 Gamma

0.547767, z4


The static colour-singlet potential
s (|q |) 4CF s (|q |) 1+ V (|q |) = - a1 + 2 q 4 + s (|q |) 4
3

s (|q |) 4 + ···

2

a2

µ2 a3 +8 C ln 2 q
2 3 A

with CA = N and CF = (N 2 - 1)/(2N ) are the eigenvalues of the quadratic Casimir operators of the adjoint and fundamental representations of the SU (N ) colour gauge group, respectively, TF = 1/2 is the index of the fundamental representation, and nl is the number of light-quark flavours.

V.A. Smirnov

PSI, November 03, 2008 ­ p.45


One loop

[W. Fischler '77; A. Billoire '80 ]

31 20 a1 = CA - TF n 9 9

l
[M. Peter'97; Y. SchrÆder'99 ]

Two loops
a2 =

4343 4 22 1798 56 2 2 +4 - + (3) CA - + (3) CA TF n 162 4 3 81 3 55 - - 16 (3) CF TF nl + 3 20 TF n 9
2 l

l

1/mq -correction

[B.A. Kniehl, A.A. Penin, V.A. Smirnov, and M. Steinhauser'02]

Three loops: a3 = + + + =? a3 is one of a few missing NNNLO ingredients of the non-relativistic dynamics near threshold.
V.A. Smirnov PSI, November 03, 2008 ­ p.46

(3) 3 a3 nl

(2) 2 a3 nl

(1) a3 nl

(0) a3


Tree and one-loop approximations:

F (a1 ,a2 ,a3 ) =

dd k (-k 2 - i0)a1 (-(q - k )2 - i0)a2 (-v · k - i0)

a3

with q · v = 0, v = (1, 0)
V.A. Smirnov PSI, November 03, 2008 ­ p.47


Two loops:

V.A. Smirnov

PSI, November 03, 2008 ­ p.48


Generation of diagrams by QGRAPH Typical diagrams in three loops:

2 2 CF ( n l T F ) 3 CF CA ( n l T F ) 2 CF ( n l T F ) 2 CF CA n l T

F

2 CF CA n l T

F

3 CF n l T

F

V.A. Smirnov

PSI, November 03, 2008 ­ p.49


Various classes of Feynman integrals with 12 indices:

V.A. Smirnov

PSI, November 03, 2008 ­ p.50


1 (-v ·k -i0)a

1 (-v ·k +i0)a

1 (-k 2 -i0)a

And 15 types of two-loop Feynman integrals with 7 indices, where one index can be shifted: ai ai + or ai ai +2
V.A. Smirnov PSI, November 03, 2008 ­ p.51


For n1 contribution: 70000 integrals of different types in l the general -gauge For example,

with the numerator chosen as (-(k - r)2 )
V.A. Smirnov

-a12
PSI, November 03, 2008 ­ p.52


Evaluating master integrals by MB. For example,

(i d/2 )3 (q 2 )3 v 2

56 135

4

+

112 135

4

+

16 2 (3) 9

+

8 (5) 3

+ O( )

(i d/2 )3 (q 2 )3 v 2

-

64 135

4

-

128 135

4

-

32 2 (3) 9

+

8 (5) 3

+ O( )

V.A. Smirnov

PSI, November 03, 2008 ­ p.53


(i d/2 ) q2 v2

3

32 135

4

-

128 135

4

+

88 2 (3) 9

+

188 (5) 3

+ O( )

Transcendentality level 5 (and, sometimes, 6) is needed. The constants of level 5 that we encounter:
(5), (4) ln 2, (2) (3), (3) ln2 2, (2) ln3 2, ln5 2, Li5 1 , Li4 1 ln 2 2 2

V.A. Smirnov

PSI, November 03, 2008 ­ p.54


a3 =

(3) 3 a3 nl

+

(2) 2 a3 nl

+

(1) a3 nl

+

(0) a3

,

[A.V. Smirnov, V.A. Smirnov, and M. Steinhauser'08 ]

a3

(3)

20 =- 9

3

3 TF ,

(2) a3

12541 368 (3) 64 4 2 CA T F , = + + 243 3 135 14002 416 (3) 2 + - CF T F , 81 3

V.A. Smirnov

PSI, November 03, 2008 ­ p.55


(1) a3

= (-709.717) C +

2 A TF

71281 + 264 (3) + 80 (5) CA CF T +- 162

F

dabcd dabcd 286 296 (3) 2 F + - 160 (5) CF TF +(-56.83(1)) F , 9 3 NA
2 Nc - 1 CF = , 2Nc

C A = Nc ,

1 TF = , 2

2 4 dabcd dabcd 18 - 6Nc + Nc F F = . 2 NA 96Nc (0)

a3 to be done.
V.A. Smirnov PSI, November 03, 2008 ­ p.56